If Sin θ = 2/3 , find the value of (1-conθ)² tan θ without using any
calculators.
S.2 maths, please help
2008-06-09 10:42 pm
回答 (3)
2008-06-09 11:35 pm
sinθ = opposite / hypotenuse, cosθ= adjacent / hypotenuse, tanθ= opposite / adjacent
adjacent^2 + 2^2 = 3^2 adjacent = √5 cosθ=√5/3 tanθ= 2/√5
留意 given sinθ (opposite 在分子) 應該知道tanθ的分子也是2
adjacent^2 + 2^2 = 3^2 adjacent = √5 cosθ=√5/3 tanθ= 2/√5
留意 given sinθ (opposite 在分子) 應該知道tanθ的分子也是2
2008-06-09 11:05 pm
sin θ = 2/3
cosθ=√(1-sinθ)=√5/3
tanθ+1=1/cosθ
tanθ=9/5-1=4/5
tanθ=2/√5
(1-cosθ) tanθ
=[1-(√5/3)]^2*2/√5
=[1-2(√5/3)+5/9]*2/√5
=(28√5/45)-4/3
cosθ=√(1-sinθ)=√5/3
tanθ+1=1/cosθ
tanθ=9/5-1=4/5
tanθ=2/√5
(1-cosθ) tanθ
=[1-(√5/3)]^2*2/√5
=[1-2(√5/3)+5/9]*2/√5
=(28√5/45)-4/3
2008-06-09 10:59 pm
cosθ = √(1 - sin2θ) = √[1-(2/3)2] = √5/3
tanθ = sinθ / cosθ = (√5/3) / (2/3) = √5/2
(1 - cosθ )2 tanθ = (1 - √5/3)2 (√5/2)
= (1 + 5/9 - 2√5/3) √5/2
= 7√5/9 - 5/3
2008-06-09 17:36:17 補充：
Revised
tanθ = sinθ / cosθ = (2/3) / (√5/3) = 2/√5
(1 - cosθ )2 tanθ = (1 - √5/3)2 (2/√5)
= (1 + 5/9 - 2√5/3) 2/√5
= 28/(9√5) - 4/3
=28√5/45 - 4/3 ( correct answer)
tanθ = sinθ / cosθ = (√5/3) / (2/3) = √5/2
(1 - cosθ )2 tanθ = (1 - √5/3)2 (√5/2)
= (1 + 5/9 - 2√5/3) √5/2
= 7√5/9 - 5/3
2008-06-09 17:36:17 補充：
Revised
tanθ = sinθ / cosθ = (2/3) / (√5/3) = 2/√5
(1 - cosθ )2 tanθ = (1 - √5/3)2 (2/√5)
= (1 + 5/9 - 2√5/3) 2/√5
= 28/(9√5) - 4/3
=28√5/45 - 4/3 ( correct answer)
收錄日期: 2021-04-13 15:40:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080609000051KK01473